Question

In a plain turning operation on steel by using cemented carbide single point cutting tool combination having a Taylor exponent is 0.25, if the cutting speed is halved, then the tool life becomes:
1. one forth
2. sixteen times
3. eight times
4. two times

Answer 1

Correct Answer - Option 2 : sixteen times

Concept:

Taylor’s Tool Life Equation based on Flank Wear:

VTn = C

Where, V = cutting speed (m/min), T = Time (min) (time taken to develop certain flank wear), n = an exponent that largely depends on tool material, C = constant based on tool and work material and cutting condition.

Calculation:

Using the Taylor’s tool life equation with exponent n = 0.25, if the cutting speed is reduced by 50%, the ratio of new tool life to original tool life is

Given: n = 0.5, V2 = 0.5 V1

V1T1n = V2T2n

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\frac{1}{n}}} = {\left( 2 \right)^{\frac{1}{{0.25}}}} = {2^4} = 16\)